Chapter 37.4 - Median of the Data

In the previous section we completed a discussion on mode. We also saw some solved examples. In this section we will discuss about median.
We have seen some basics about median in part II (Details here). 
It is the middle most observation in a data. We saw the following steps:
• Arrange all data values of the observations in ascending order
• Let the total number of observations be 'n'
• If n is odd, the median is the observation at the position given by: ⁽ⁿ⁺¹⁾⁄₂
• If n is even, the median is the average of the following two observations:
    ♦ observation at the (ⁿ⁄₂)^th position
    ♦ observation at the (ⁿ⁄₂ + 1)^th position

---

Let us see a new example:
Example 8:
The following table 37.34 shows the marks (out of 50) obtained by 100 students in a test

Table.37.34

We have to find the median mark
Solution:
• How does this problem differ from the previous 'problems on median' that we saw in part II?
Ans: In this problem, each mark has a frequency. Because, each mark (except 43) occur more than once
    ♦ 20 occurs 6 times
    ♦ 29 occurs 28 times
So on . . .
• Then how can we arrange the marks in ascending or descending order?
Ans: We will develop a new method for such problems. The steps are as follows:
1. Arrange the data in ascending order, without considering the frequencies. This is shown in table 37.35 below:

Table.37.35

2. In this problem, n = 100. It is even. So median will be the average of the following two observations:
    ♦ observation at the (ⁿ⁄₂)^th position
    ♦ observation at the (ⁿ⁄₂ + 1)^th position
3. So we want these:
• observation at the (ⁿ⁄₂)^th position
    ♦ That is., the observation at the 50^th position 
• observation at the (ⁿ⁄₂ + 1) position. 
    ♦ That is., the observation at the 51^st position 
4. Obviously, we cannot obtain them directly from the table 37.34 given in the question or from the table 37.35 above. We will use a special method:
• The data in table 37.35 is arranged in ascending order. 
Based on the table, let us try to answer some questions:
(i) How many students obtained at least 20 marks?
Ans: Considering the second row of the table, we can easily write:
• No. of students who obtained at least 20 marks is 6
(ii) How many students obtained at least 25 marks?
Ans: The number of students who obtained 20 marks will also be included when we consider 'at least' 25.
• So the no. of students who obtained at least 25 marks is (6+20) = 26
(iii) How many students obtained at least 28 marks?
Ans: The number of students who obtained at least 25 marks will also be included when we consider 'at least' 28.
• So the no. of students who obtained at least 28 marks is (26+24) = 50
5. We can go on answering like this till we reach the end of the table. But once we understand the pattern, we can quickly fill up the values
• This new information can be written as cumulative frequency. We will add it to the existing table. The new table 37.36 is given below:

Table.37.36

• Note how easy it is to fill up the 'cumulative frequency column':
To fill up any one row in the 'cumulative frequency column', do these steps:
(i) Take the value coming in the same row in the 'frequency column'
(ii) Take the value coming in the row just above in the 'cumulative frequency column'
(iii) Add (i) and (ii)
6. Now, if we write all the observations in a single line, it will look as in the table.37.37 below:

Table.37.37

• The first 6 observations will be all 20
• The 7^th, 8^th, 9^th, . . . , 26^th observations will all be 25
• The 27^th, 28^th, 29^th, . . . , 50^th observations will all be 28
so on . . . 
7. So the 50^th observation is 28
And the 51^st observation is 29
• These are the observations that we want.
• But we do not need to form a table like the table 37.37 above. We will be able to find them easily from the 'cumulative frequency column' in table 37.36
8. Thus from step (3) above, we get:
Median = ^(28+29)⁄₂ = ⁵⁷⁄₂ = 28.5
9. So we can write a conclusion:
• If the 100 marks are arranged in ascending order, 28.5 will come in the middle
• So about 50 students obtained marks less than 28.5
• The other 50 students obtained marks more than 28.5
■ The table 37.36 above is called the Cumulative frequency table

---

The example above shows the procedure to find median of observations in an 'ungrouped frequency distribution table'
So next we will see the procedure for 'grouped frequency distribution tables':
Example 9:
The following table 37.38 shows the marks (out of 50) obtained by 53 students in a test

Table.37.38

We have to find the median mark
Solution:
We will write the steps:
1. Based on the table 37.38, let us try to answer some questions:
(i) How many students have obtained marks less than 10?
Ans: Considering the class '0 -10', we can say:
• 5 students obtained marks less than 10
• Note that, we use 'less than' in this problem
• In the previous problem it was 'at least'
    ♦ If we use 'at least' in this problem, the mark '10' will be included
    ♦ We know that, in the class '0 - 10', the mark '10' will not be considered. 
    ♦ It will be considered only in the next higher class, which in this problem is '10 - 20'
(ii) How many students have obtained marks less than 20?
Ans: The number of students who obtained marks less than 10 will also be included when we consider 'less than' 20.
• So the no. of students who obtained marks less than 20 is (5+3) = 8
(iii) How many students have obtained marks less than 30?
Ans: The number of students who obtained marks less than 20 will also be included when we consider 'less than' 30.
• So the no. of students who obtained marks less than 30 is (8+4) = 12
• We can go on answering like this till we reach the end of the table. But once we understand the pattern, we can quickly fill up the values

• This new information can be written as cumulative frequency. We will add it to the existing table. The new table 37.39 is given below:

Table.37.39

• Note how easy it is to fill up the 'cumulative frequency column':
To fill up any one row in the 'cumulative frequency column', do these steps:
(i) Take the value coming in the same row in the 'frequency column'
(ii) Take the value coming in the row just above in the 'cumulative frequency column'
(iii) Add (i) and (ii)
■ The distribution given above is called the cumulative frequency distribution of the less than type.
• Here 10, 20, 30, . . . 100, are the upper limits of the respective class intervals.
2. So let us see what 'cumulative frequency distribution of the more than type' is:
(i) How many students have obtained marks more than 0?
Ans: We can say that all the 53 students obtained marks more than 0
(ii) How many students have obtained marks more than 10?
Ans: Considering the class '0 -10', we can say:
• 5 students obtained marks less than 10
So the number of students who obtained marks more than 10 is (53-5) = 48
(iii) How many students have obtained marks more than 20?
Ans: From (i) we have: Number of students who obtained marks more than 10 = 48
Considering the class '10 - 20', we can say:
• 3 students among the above 48, obtained marks less than 20
• So the number of students who obtained marks more than 20 is (48-3) = 45
(iv) How many students have obtained marks more than 30?
Ans: From (ii) we have: Number of students who obtained marks more than 20 = 45
Considering the class '20 - 30', we can say:
• 4 students among the above 45, obtained marks less than 30
• So the number of students who obtained marks more than 30 is (45-4) = 41
• We can go on answering like this till we reach the end of the table. But once we understand the pattern, we can quickly fill up the values

• This new information can be written as cumulative frequency. We will add it to the existing table. The new table 37.40 is given below:

Table.37.40

• Note how easy it is to fill up the 'cumulative frequency column':
To fill up any one row in the 'cumulative frequency column', do these steps:
(i) Take the value coming in the same row in the 'frequency column'
(ii) Take the value coming in the row just above in the 'cumulative frequency column'
(iii) Subtract (i) from (ii)
■ The distribution given above is called the cumulative frequency distribution of the more than type.

• Here 10, 20, 30, . . . 90, are the lower limits of the respective class intervals.
3. So we have obtained two types of cumulative frequency distribution tables. We can use any one of them to find the median. Let us try:
• We will use the less than type. So we will use the table 37.39. The steps are as follows:
(i) In this problem, the total number of observations 'n' = 53. This is an odd number.
So ⁽ⁿ⁺¹⁾⁄₂ = ⁵⁴⁄₂ = 27
• In the previous example 8, we could easily find the observation corresponding to ⁿ⁄₂. This is because, the data given to us in that question is an 'ungrouped frequency distribution table' 
• In such a table, we can find the position of each observation (when arranged in ascending order) from the 'cumulative frequency column'
• But in our present problem, we are given a 'grouped frequency distribution table'. So the actual observations are hidden inside various groups. Even then we will try.
• Since the number of observations is 53, which is odd, the middle observation is the observation at the 27^th position (when arranged in ascending order)
• So let us try to find the observation at the 27^th position. We will try to find it from table 37.39 above. We will travel through the table from top to bottom and examine each class carefully.
• Any 'grouped frequency distribution table' will already have an 'ascending order arrangement' because, the classes are in the increasing order. In our problem, it is: 0-10, 10-20, 20-30 so no...
• Now, the first 5 observations go inside the first class interval 0-10
    ♦ That means, the 5^th observation goes inside the first class interval
    ♦ We are not concerned because, we need the 27^th observation. Not the 5^th.
• The first 8 observations go inside the first and second class intervals 0-10 and 10-20
    ♦ That means, the 8^th observation goes inside the first and second classes taken together
    ♦ We are not concerned because, we need the 27^th observation. Not the 8^th.
• The first 12 observations go inside the first, second and third class intervals 0-10, 10-20 and 20-30
    ♦ That means, the 12^th observation goes inside the first, second and third classes taken together
    ♦ We are not concerned because, we need the 27^th observation. Not the 12^th.
• Continuing in this way, we get:
    ♦ 22^nd observation falls within 50-60
    ♦ 29^th observation falls within 60-70
• There is no sign of 27. 
• The 27^th observation is some where in between the 22^nd and 29^th.
    ♦ But there are no classes in between 50-60 and 60-70
• So the 27^th observation has to be within one among the two: 50-60 or 60-70
But which one?
• 50-60 cannot take in any one higher than 22
• So it has to be within the 60-70
■ So we can conclude:
The 27^th observation is inside the 60-70 class

---

To find the appropriate class, we do not need a detailed analysis as given above. Once we understand the basics, we can straight away take out the class whose cumulative frequency is closest to and just greater than the 'position we are seeking'.
• In our present problem, we are seeking the 27^th position
• The cumulative frequency closest to and just greater than 27 is 29
• So take out 60-70

---

• The class thus taken out is called the median class
4. Our median class contains 7 observations. 
• They may be any value equal to or greater than 60 and less than 70
• The median is one of those 7 observations. Our next aim is to find it
■ The median is calculated using the following formula:

Where:
l = lower limit of the median class
n = number of observations
cf = cumulative frequency of the class preceding the median class
f = frequency of the median class
h = width of the class interval (assuming all classes are of the same width)
• Substituting the values, we get:
Numerator = (ⁿ⁄₂ - cf) = (⁵³⁄₂ - 22)  = (26.5-22) = 4.5
Thus median = 60 + (^4.5⁄₇)×10 = 60 + 6.4 = 66.4
We can write the conclusion:
If we arrange all the 53 marks in ascending order, the mark 66.4 will come in the middle. 
• 27 marks will lie on the left of 66.4
    ♦ All those marks will be less than 66.4
• 27 marks will lie on the right of 66.4
    ♦ All those marks will be greater than 66.4

Example 10:
A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained:

Table.37.41

Find the median height
Solution:
1. The given data is in the form of a cumulative frequency table. 
We have to convert it into a 'grouped frequency distribution table'. Let us write the steps:
• No. of heights less than 140 is 4
• No. of heights less than 145 is 11
• This 11 will include the heights less than 140 also
2. From the above three information, we can find the no. of heights satisfying the following two conditions:
    ♦ No. of heights greater than or equal to 140
    ♦ No. of heights less than 145
• Obviously, the no. of heights satisfying the two conditions is: (11-4) = 7
• So 7 heights satisfy the condition: 140 ≤ height < 145
• Those 7 heights will obviously fall in the class: 140-145
• In other words, the frequency of the class 140-145 is 7
3. So we separated out 7 heights from 11
What about the remaining 4?
• They satisfy the condition: 'height < 140'
• Those 4 heights will obviously fall in the class: k-140
    ♦ Where 'k' is any number less than 140. We will give it a value 135
• So the frequency of the class 135-140 is 4
4. So we found two classes: 
    ♦ class 135-140 with frequency 4
    ♦ class 140-145 with frequency 7
Now we consider the next cumulative frequency, which is 29:
• No. of heights less than 145 is 11
• No. of heights less than 150 is 29
• This 29 will include the heights less than 145 also
• From the above three information, we can find the no. of heights satisfying the following two conditions:
    ♦ No. of heights greater than or equal to 145
    ♦ No. of heights less than 150
• Obviously, the no. of heights satisfying the two conditions is: (29-11) = 18
• So 18 heights satisfy the condition: 145 ≤ height < 150
• Those 18 heights will obviously fall in the class: 145-150
• In other words, the frequency of the class 145-150 is 18
5. Thus so far, we found three classes: 
    ♦ class 135-140 with frequency 4
    ♦ class 140-145 with frequency 7
    ♦ class 145-150 with frequency 18
Now we consider the next cumulative frequency, which is 40:
• No. of heights less than 150 is 29
• No. of heights less than 155 is 40
• This 40 will include the heights less than 150 also
• From the above three information, we can find the no. of heights satisfying the following two conditions:
    ♦ No. of heights greater than or equal to 150
    ♦ No. of heights less than 155
• Obviously, the no. of heights satisfying the two conditions is: (40-29) = 11
• So 11 heights satisfy the condition: 150 ≤ height < 155
• Those 11 heights will obviously fall in the class: 150-155
• In other words, the frequency of the class 150-155 is 11
• Thus so far, we found four classes: 
    ♦ class 135-140 with frequency 4
    ♦ class 140-145 with frequency 7
    ♦ class 145-150 with frequency 18
    ♦ class 150-155 with frequency 11
■ We can go on like this until we reach the end of the table. However, once we understand the pattern, the frequencies of each class can be easily calculated. The resulting table is shown below:

Table.37.42

• Note how easy it is to fill up the 'frequency column':
To fill up any one row in the 'frequency column', do these steps:
(i) Take the value coming in the same row in the 'cumulative frequency column'
(ii) Take the value coming in the row just above in the 'cumulative frequency column'
(iii) Subtract (ii) from (i)
6. Now we can begin the calculations for the median:
• In this problem, the total number of observations 'n' = 51. This is an odd number.
So ⁽ⁿ⁺¹⁾⁄₂ = ⁵²⁄₂ = 26
• The cumulative frequency closest to 26 and just greater than 26 is 29
• The class corresponding to the cumulative frequency 29 is 145-150
• So 145-150 is the median class. It's frequency = 18
■ The median is calculated using the following formula:

Where:
l = lower limit of the median class = 145
n = number of observations = 51
cf = cumulative frequency of the class preceding the median class = 11
f = frequency of the median class = 18
h = width of the class interval (assuming all classes are of the same width) = 5
• Substituting the values, we get:
• Numerator = (ⁿ⁄₂ - cf) = (⁵¹⁄₂ - 11)  = (25.5-11) = 14.5
• Thus median = 145 + (^14.5⁄₁₈)×5 = 145 + 4.03 = 149.03
7. We can write the conclusion:
If all the 51 girls stand in ascending order of heights, the girl with height 149.03 will come in the middle. 
• 25 girls will stand on the left of this girl
    ♦ Heights of all those girls will be less than 149.03 cm
• 25 girls will stand on the right of this girl
    ♦ Heights of all those girls will be greater than 149.03 cm

---

In the next section, we will see another example.

PREVIOUS      CONTENTS       NEXT

 
                        Copyright©2018 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 28.2- Tossing two coins simultaneously

In the previous section we saw the results when the number of times a die is rolled is increased. In this section we will see such experiments related to 'tossing two coins simultaneously'.

Let us do an experiment. We will call it experiment IIIA:
1. Take two coins. 
(i) Toss them simultaneously. The reading may be any one of the following:
HH, TT and HT
Note that, in the result HT, it does not matter which coin gives H and which one T
(ii) What ever be the reading, note that reading on a piece of paper. 
• (i) and (ii) constitutes one cycle of our experiment.
2. Repeat the cycle 10 times. 
3. In the note book, tabulate the readings as shown in table 28.5 below 
Table 28.5

Number of times the two coins are tossed	Number of times no heads comes up (TT)	Number of times one head comes up (HT)	Number of times two heads come up (HH)
10	2	6	1

4. Determine the following ratios:
• ^{Number of times no head turned up}⁄_{Total number of times the two coins are tossed}
• ^{Number of times one head turned up}⁄_{Total number of times the two coins are tossed}
• ^{Number of times two heads turned up}⁄_{Total number of times the two coins are tossed}
• In our present case:
    ♦ the 1^st ratio is ²⁄₁₀ = 0.2
    ♦ the 2^nd ratio is ⁶⁄₁₀ = 0.6
    ♦ the 3^rd ratio is ¹⁄₁₀ = 0.1
5. Once the ratios in (4) are determined, the experiment IIIA is complete

---

But our work is not over

■ Repeat the above experiment. We will call it experiment IIIB.
• For this experiment IIIB, the number of cycles in (2) must be 20
• Let the readings be as shown below:

Number of times the two coins are tossed	Number of times no heads comes up (TT)	Number of times one head comes up (HT)	Number of times two heads come up (HH)
20	5	11	4

• In this case:
    ♦ the 1^st ratio is ⁵⁄₂₀ = 0.25
    ♦ the 2^nd ratio is ¹¹⁄₂₀ = 0.55
    ♦ the 3^rd ratio is ⁴⁄₂₀ = 0.2
• When the ratios in (4) are determined, the experiment IIIB is over
■ Once again repeat the experiment. We will call it experiment IIIC
• For this experiment IIIC, the number of cycles in (2) must be 30
• Let the readings be as shown below:

Number of times the two coins are tossed	Number of times no heads comes up (TT)	Number of times one head comes up (HT)	Number of times two heads come up (HH)
30	8	17	5

• In this case:
    ♦ the 1^st ratio is ⁸⁄₃₀ = 0.267
    ♦ the 2^nd ratio is ¹⁷⁄₃₀ = 0.567
    ♦ the 3^rd ratio is ⁵⁄₃₀ = 0.167
• When the ratios in (4) are determined, the experiment IIIC is over

---

So we did the same experiment 3 times. Before proceeding further, we will discuss the importance of the six ratios:

1. We know that, the probability of obtaining TT, HT and HH when tossing two coins simultaneously are 0.25, 0.5 and 0.25 respectively

2. But these are theoretical values. If they are always obtained in the real life also, we will get results such as these:

• Toss two coins simultaneously 12 times
    ♦ TT will be obtained 3 times
    ♦ HT will be obtained 6 times
    ♦ HH will be obtained 3 times
• Toss two coins simultaneously 16 times
    ♦ TT will be obtained 4 times
    ♦ HT will be obtained 8 times
    ♦ HH will be obtained 4 times

3. But we never get such exact values.
4. However, as the number of trials increase, each of the 3 ratios become closer and closer to their respective values 0.25, 0.5 and 0.25

5. If, instead of 10,20 or 30 times, if we toss it for a 'very large number of times, n', then:
• ^{Number of times TT is obtained}⁄_n = 0.25
• ^{Number of times HT is obtained}⁄_n = 0.5
• ^{Number of times HH is obtained}⁄_n = 0.25

We are trying to prove this using our present experiments

---

• From the three experiments, we have three sets of ratios. One set from each experiment

• Each set has six ratios:
    ♦ ^{Number of times TT is obtained}⁄_n
    ♦ ^{Number of times HT is obtained}⁄_n
    ♦ ^{Number of times HH is obtained}⁄_n
■ Let us now analyse the ratios:
1^st ratio when number of trials is 10 = 0.2
1^st ratio when number of trials is 20 = 0.25
1^st ratio when number of trials is 30 = 0.267
■ As the number of trial increases, the 1^st ratio gets closer and closer to 0.25
Consider the 2^nd ratio:
2^nd ratio when number of trials is 10 = 0.6
2^nd ratio when number of trials is 20 = 0.55
2^nd ratio when number of trials is 30 = 0.567
■ As the number of trial increases, the 2^nd ratio gets closer and closer to 0.5
Consider the 3^rd ratio:
3^rd ratio when number of trials is 10 = 0.1
3^rd ratio when number of trials is 20 = 0.2
3^rd ratio when number of trials is 30 = 0.167
■ As the number of trial increases, the 3^rd ratio gets closer and closer to 0.25

---

• We can increase the 'number of trials' to a 'considerably large value'  by using groups as we saw in the previous sections. Students may try it themselves.
• It will become clear that the ratios get closer and closer to 0.25, 0.5 and 0.25

---

In the next section, we will see how the above findings can be used to solve practical problems.  

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 28.1 - Increasing number of Rolls for a Die

In the previous section we saw the results when the number of times a coin is tossed is increased. In this section we will see such experiments related to 'rolling a die'.

Let us do an experiment. We will call it experiment IIA:
1. Take a die. 
(i) Roll it once. The reading may be any one of the following:
1, 2, 3, 4, 5 or 6
(ii) What ever be the reading, note that reading on a piece of paper. 
• (i) and (ii) constitutes one cycle of our experiment.
2. Repeat the cycle 20 times. 
3. In the note book, tabulate the readings as shown in table 28.3 below 
Table 28.3

Number of times a die is thrown	Number of times these scores turn up
	1	2	3	4	5	6
20	7	1	1	3	5	3

4. Determine the following ratios:
• ^{Number of times 1 turned up}⁄_{Total number of times the die is rolled} 
• ^{Number of times 2 turned up}⁄_{Total number of times the die is rolled} 
• ^{Number of times 3 turned up}⁄_{Total number of times the die is rolled} 

• - - - 
• - - - 
• ^{Number of times 6 turned up}⁄_{Total number of times the die is rolled} 
• In our present case:
    ♦ the 1^st ratio is ⁷⁄₂₀ = 0.35
    ♦ the 2^nd ratio is ¹⁄₂₀ = 0.05
    ♦ the 3^rd ratio is ¹⁄₂₀ = 0.05
    ♦ the 4^th ratio is ³⁄₂₀ = 0.15
    ♦ the 5^th ratio is ⁵⁄₂₀ = 0.25
    ♦ the 6^th ratio is ³⁄₂₀ = 0.15
5. Once the ratios in (4) are determined, the experiment IIA is complete

---

But our work is not over

■ Repeat the above experiment. We will call it experiment IIB.
• For this experiment IIB, the number of cycles in (2) must be 40
• Let the readings be as shown below:

Number of times a die is thrown	Number of times these scores turn up
	1	2	3	4	5	6
40	4	8	9	4	9	6

• In this case:
    ♦ the 1^st ratio is ⁴⁄₄₀ = 0.1
    ♦ the 2^nd ratio is ⁸⁄₄₀ = 0.2
    ♦ the 3^rd ratio is ⁹⁄₄₀ = 0.225
    ♦ the 4^th ratio is ⁴⁄₄₀ = 0.1
    ♦ the 5^th ratio is ⁹⁄₄₀ = 0.225
    ♦ the 6^th ratio is ⁶⁄₄₀ = 0.15
• When the ratios in (4) are determined, the experiment IIB is over
■ Once again repeat the experiment. We will call it experiment IIC
• For this experiment IIC, the number of cycles in (2) must be 60 
• Let the readings be as shown below:

Number of times a die is thrown	Number of times these scores turn up
	1	2	3	4	5	6
60	9	10	8	10	12	11

• In this case:
    ♦ the 1^st ratio is ⁹⁄₆₀ = 0.15
    ♦ the 2^nd ratio is ¹⁰⁄₆₀ = 0.167
    ♦ the 3^rd ratio is ⁸⁄₆₀ = 0.133
    ♦ the 4^th ratio is ¹⁰⁄₆₀ = 0.167
    ♦ the 5^th ratio is ¹²⁄₆₀ = 0.2
    ♦ the 6^th ratio is ¹¹⁄₆₀ = 0.183
• When the ratios in (4) are determined, the experiment IIC is over

---

So we did the same experiment 3 times. Before proceeding further, we will discuss the importance of the six ratios:

1. We know that, the probability of obtaining 1, 2, 3, 4, 5 or 6 when rolling a die once is ¹⁄₆

• Each of the numbers 1, 2, 3, 4, 5 and 6 has the same probability ¹⁄₆.

2. But this '¹⁄₆' is a theoretical value. If ¹⁄₆ is always obtained in the real life also, we will get results such as these:

• Roll the die 6 times
    ♦ 1 will be obtained once
    ♦ 2 will be obtained once
    ♦ 3 will be obtained once
    ♦ - - -
    ♦ - - - 
    ♦ 6 will be obtained once
• Roll the die 18 times
    ♦ 1 will be obtained thrice
    ♦ 2 will be obtained thrice
    ♦ 3 will be obtained thrice
    ♦ - - -
    ♦ - - - 
    ♦ 6 will be obtained thrice

3. But we never get such exact values.
4. However, as the number of trials increase, each of the 6 ratios become closer and closer to ¹⁄₆

5. If, instead of 20,40 or 60 times, if we toss it for a 'very large number of times', then: 

• No. of times 1 is obtained = No. of times 2 is obtained = No. of times 3 is obtained = No. of times 4 is obtained = No. of times 5 is obtained = No. of times 6 is obtained. = k

• That is., 'the equality of probability while rolling a die' will become clear.
• Let the 'very large number of times' for which the die is rolled be 'n'
• Then ^k⁄_n will be equal to ¹⁄₆.

5. We are trying to prove this using our present experiments

---

• From the three experiments, we have three sets of ratios. One set from each experiment

• Each set has six ratios:
    ♦ ^{Number of times 1 turned up}⁄_{Total number of times the die is rolled} 
    ♦ ^{Number of times 2 turned up}⁄_{Total number of times the die is rolled} 
    ♦ ^{Number of times 3 turned up}⁄_{Total number of times the die is rolled} 
    ♦ - - - 
    ♦ - - - 
    ♦ ^{Number of times 6 turned up}⁄_{Total number of times the die is rolled} 
■ Let us now analyse the ratios:
1^st ratio when number of trials is 20 = 0.35
1^st ratio when number of trials is 40 = 0.1
1^st ratio when number of trials is 60 = 0.15
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 1^st ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 1^st ratio gets closer and closer to ¹⁄₆.
Consider the 2^nd ratio:
2^nd ratio when number of trials is 20 = 0.05
2^nd ratio when number of trials is 40 = 0.2
2^nd ratio when number of trials is 60 = 0.167
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 2^nd ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 2^nd ratio gets closer and closer to ¹⁄₆.
Consider the 3^rd ratio:
3^rd ratio when number of trials is 20 = 0.05
3^rd ratio when number of trials is 40 = 0.225
3^rd ratio when number of trials is 60 = 0.133
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 3^rd ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 3^rd ratio gets closer and closer to ¹⁄₆.
Consider the 4^th ratio:
4^th ratio when number of trials is 20 = 0.15
4^th ratio when number of trials is 40 = 0.1
4^th ratio when number of trials is 60 = 0.167
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 4^th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 4^th ratio gets closer and closer to ¹⁄₆.
Consider the 5^th ratio:
5^th ratio when number of trials is 20 = 0.25
5^th ratio when number of trials is 40 = 0.225
5^th ratio when number of trials is 60 = 0.2
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 5^th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 5^th ratio gets closer and closer to ¹⁄₆.
Consider the 6^th ratio:
6^th ratio when number of trials is 20 = 0.15
6^th ratio when number of trials is 40 = 0.15
6^th ratio when number of trials is 60 = 0.183
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 6^th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 6^th ratio gets closer and closer to ¹⁄₆.

---

So all the ratios are getting closer and closer to ¹⁄₆. When the number of trials become very large, each of the ratios will become equal to ¹⁄₆.

---

• We can increase the 'number of trials' to a 'considerably large value' in another way. We will do it as a new experiment. We will call it: Experiment IIC
1. Divide the class into a number of groups. Each group must have three students
In our present case, let there be 15 groups, each with 3 students
2. Consider any one group
(i) A student in that group rolls the die once. The reading may be 1, 2, 3, 4, 5 or 6
(ii) Whatever be the reading, the other two students must check it and record it
• (i) and (ii) constitutes one cycle of our experiment.
3. Repeat the cycle 20 times. When the 20 cycles are completed, the task of this group of students is complete.
4. Now the die can be given to another group. They can do the 20 cycles.
5. After the 20 cycles, the die can be given to yet another group
6. So, when all the 15 groups have completed their tasks, the die will be rolled 15×20 = 300 times. 
• This 300 is a significantly large number. So we achieve a 'large task of rolling a die 300 times'. We achieve it by dividing the task among 15 groups.
7. If there are 15 dice, all the groups can perform the task simultaneously. This will save time. 
• There is only one condition: All the 15 dice must be identical.
If this condition is satisfied, it can be considered that a single student did all the 300 rolls.
8. When all the groups have completed their tasks, we can proceed to do the calculations. We have to combine the results in such a way that, one student did all the 300 rolls. 
9. For that, do the tabulation as shown in the table 28.4 below. 
• In the table, only 1 and 4 is given. This is to keep the size of the table small. Students may use a large sheet of paper and include all the numbers from 1 to 6. 
• The explanation about 'how this table is formed' is given below the table.

Table 28.4

Group	Number of 1	Number of 4	A	B
(1)	(2)	(3)	(4)	(5)
1	3	4	3⁄20 = 0.15	4⁄20 = 0.2
2	2	1	(2+3)⁄(20+20) = 5⁄40 = 0.125	(1+4)⁄(20+20) = 5⁄40 = 0.125
3	2	5	(2+5)⁄(20+40) = 7⁄60 = 0.116	(5+5)⁄(20+40) = 10⁄60 = 0.16
_	_	_	_	_
_	_	_	_	_
15	_	_	_	_

■ Consider the row corresponding to group 1. 
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
    ♦ In the 4^th column, we have a ratio ³⁄₂₀. 
    ♦ Obviously, it is the ratio: ^{Number of times 1 comes up}⁄_{Total number of times the die is rolled}
    ♦ In the 5^th column, we have a ratio ⁴⁄₂₀. 
    ♦ Obviously, it is the ratio: ^{Number of times 4 comes up}⁄_{Total number of times the die is rolled}
■ Consider the row corresponding to group 2.  
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
    ♦ In the 4^th column, we have a ratio ⁵⁄₄₀.
    ♦ The numerator is (2+3). So it is the total number of 1 obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 40 rolls, and obtaining 1 five times'.
    ♦ So in the denominator we have 40  
    ♦ In the 5^th column, we have a ratio ⁵⁄₄₀.
    ♦ The numerator is (1+4). So it is the total number of 4 obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 40 rolls, and obtaining 4 five times'.
    ♦ So in the denominator we have 40  
■ Consider the row corresponding to group 3.  
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
    ♦ In the 4^th column, we have a ratio ⁷⁄₆₀.
    ♦ The numerator is (2+5). So it is the total number of 1 obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 60 rolls, and obtaining 1 seven times'.
    ♦ So in the denominator we have 60  
    ♦ In the 5^th column, we have a ratio ¹⁰⁄₆₀.
    ♦ The numerator is (5+5). So it is the total number of 4 obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 60 rolls, and obtaining 4 ten times'.
    ♦ So in the denominator we have 60

---

■ In this way, we can fill up the rows of all the 15 groups, and find all the required ratios.

■ Note: As the number of '1' and '4' in each row is added to the 'total values up to the previous row', the word cumulative can be used.

• So the heading of the 4^th column (which is written as A in the table) is:

^{Cumulative number of 1}⁄_{Total number of times the die is rolled}

• Similarly, the heading of the 5^th column (which is written as B in the table) is:

^{Cumulative number of 4}⁄_{Total number of times the die is rolled}

---

Once the ratios are found out, we can analyse them:

■ Group 1:

• Number of rolls = 20
Ratio for '1' = 0.15
Ratio for '4' = 0.2

■ Groups 1 and 2:
• Number of rolls = 40
Ratio for '1' = 0.125
Ratio for '4' = 0.125

■ Groups 1, 2 and 3:

• Number of rolls = 60
Ratio for '1' = 0.116
Ratio for '4' = 0.16
■ So the ratio for '1' takes the values 0.15, 0.125, 0.116, . . .
It is getting closer and closer to ¹⁄₆.
■ The ratio for '4' takes the values 0.2, 0.125, 0.16
It is getting closer and closer to ¹⁄₆.
• If we had made the above table 28.4 on a large sheet of paper, that included numbers 2, 3, 5 and 6 also, we would get the same result for them also.  

■ In this experiment we made 300 trials. This '300' is significantly large but not even close to 'a very large number n'. 
• A very large number is like 'infinity'
• At such a very large number, all the ratios will become equal to ¹⁄₆.

---

In the next section, we will see the above procedure when 'two coins are tossed simultaneously'.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Maths lessons. blogspot.in - All Rights Reserved